ASSIGNMENT 8 - PROBLEM 12

Given an acute triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. Prove:

PART ONE - CONSTRUCTION:

In this part, I will just refer to orthocenter file of assignment five.

Let's rename the points so that H is the orthocenter of triangle ABC and D, E, and F are the feet of the perpendiculars from A, B, and C respectfully.

After constructing triangle ABC with the requested conditions, in PART - II below I will prove the first equality and in PART - III, I will prove the second equality.

PART TWO - PROVE

Before proving this equality, I would ask my students to play with it using GSP. In this way they can see that what they are trying to prove is really true. Click the figure below to play with the GSP file:

I will prove this equality via area modeling. Therefore, first let's see that the area of ABC is equal to the sum of the areas of AHB, BHC, and AHC:

Here is the proof via area modeling:

Step One: Simplify notation first. Let's define the area of triangle ABC to be a.

Area Triangle AHB + Area Triangle BHC + Area Triangle AHC = Area Triangle ABC

Step Three: Divide each side of the equation by the Area of Traingle ABC, namely by a.

(1/a)(Area Triangle AHB + Area Triangle BHC + Area Triangle AHC ) = 1

Step Four: Use distributive property of multiplication over addition:

(1/a)(Area Triangle AHB )+ (1/a)(Area Triangle BHC) + (1/a)(Area Triangle AHC ) = 1

Step Five: Write the areas of each triangle in terms of theirs bases and heights: AREA = 0.5 x BASE x HEIGHT

(1/a) (0.5) (AB) (FH) + (1/a) (0.5) (BC) (DH) + (1/a) (0.5) (AC) (EH) = 1

Step Six: We are close now to what we want to prove. It just remains to divide each one of the three terms on the left hand side by the appropriate a of the Area of Triangle ABC.

Step Seven: Area of Triangle ABC, namely a, can be expressed in three different ways:

a = (0.5) (AB) (FC) , OR , a = (0.5) (BC) (AD) , OR , a = (0.5) (AC) (BE). The following little table summarizes this:

 FIRST FORM SECOND FORM THIRD FORM a = (0.5) (AB) (FC) a = (0.5) (BC) (AD) a = (0.5) (AC) (BE)

Step Eight: If we use the first form of a with the first term on the left hand side of the equation in Step Five above, then it simplifies to AH / FC

Step Nine: If we use the second form of a with the second term on the left hand side of the equation in Step Five above, then it simplifies to DH / AD

Step Eight: If we use the third form of a with the third term on the left hand side of the equation in Step Five above, it simplifies to EH / BE

Step Ten: Rewriting the equation of Step Five in simplified form, we get (FH / FC) + (DH / AD) + (EH / BE) = 1 and this completes the proof.

PART THREE - PROVE

Before proving this equality, I would ask my students to play with it using GSP. In this way they can see that what they are trying to prove is really true. Click the figure below to play with the GSP file:

I will prove this equality by using the proof of the first identity I proved in PART - II above:

Step One: DH / AD + EH / BE + FH / CF= 1

Step Two: Comparing this expression with what we want to prove, we see that the denominator of each term is the same.

Step Three: Express DH in terms of AH. Namely, DH = AD - AH.

Step Four: Express EH in terms of BH. Namely, EH = BE - BH.

Step Five: Express FH in terms of CH. Namely, FH = CF - CH.

Step Six: Now plug these in the equation of Step One:

(AD - AH) / AD + (BE - BH) / BE + (CF - CH) / CF= 1

Step Seven: This time, use distributive property of division over subtraction:

[(AD / AD) - (AH / AD)] + [(BE / BE) - (BH / BE)] + [(CF / CF) - (CH / CF)] = 1

Step Eight: Simplify to get

[ 1 - (AH / AD)] + [ 1 - (BH / BE)] + [ 1 - (CH / CF)] = 1

Step Nine: Rewrite this as

3 - [ (AH / AD) + (BH / BE) + (CH / CF) ] = 1

Step Ten: Solve for the term in the square bracket

[ (AH / AD) + (BH / BE) + (CH / CF) ] = 2

and this completes the proof.

PART FOUR - WHAT HAPPENS IF ABC IS AN OBTUSE TRIANGLE?

What we proved is valid only when the orthocenter H is inside the triangle ABC, in other words, for acute triangles. Indeed, we can see this by clicking the figure below that comprises the two cases together:

PART FIVE - ORTHOPATH REVIEW:

In Assignment 4, as an application of parabola, we learned that the trajectory of the orthocenter of a triangle is a parabola when one of the bases is fixed and the vertex opposite to this base is animated along aline that is parallel to this base. In our case, if we fix the base BC while animating the point A along an invisible line parallel to BC, we realize that the orthocenter H traces a parabola. Click the figure below to download this: